How to set a location for a popup window in windows form?

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In a form I have a datagridview which has a contents of my xml file. I am opening a popup settings window to on cell right click. Here popup setting window is a another form.

I have written a code for opening a popup setting window form. but it is showing on the topmost corner of the screen. how to set the location of the popup setting window form below the currentcell in a datagridview.

What I have tried:

C#

private void dataGridView1_CellMouseUp(object sender, DataGridViewCellMouseEventArgs e)
        {
 if (e.Button == MouseButtons.Right)
  {
private frmDGVColList objform = new frmDGVColList();
objform.StartPosition = FormStartPosition.Manual;
objform.Location = dataGridView1.PointToScreen(dataGridView1.GetCellDisplayRectangle(e.ColumnIndex, e.RowIndex, false).Location);
objform.Show(this.dataGridView1);
}
}

Posted 9-Apr-18 2:22am

Poornima Santhosh

Updated 9-Apr-18 2:44am

CHill60

v2

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RickZeeland · Answer 1 · 2018-04-09T02:45:00

Solution 1

This seems to be what you are looking for: c# - Open a popup from just below of current selected cell of datagrid view - Stack Overflow[^]

Posted 9-Apr-18 2:45am

RickZeeland

Comments

Poornima Santhosh 9-Apr-18 8:48am

I have tried its not working

RickZeeland 9-Apr-18 8:51am

Did you also modify your popup form like this ?

private Point loc;
public Form2(Point location)
{
InitializeComponent();
loc = location;
}

private void Form2_Load(object sender, EventArgs e)
{
this.SetDesktopLocation(loc.X, loc.Y);
}